The Magnetic pressure Exerted upon a Magnetic Dipole

We now start our study of magnetism, and, analogous come the means in i m sorry we started our research of electricity, we start by pointing out the effect of a provided magnetic field without very first explaining how such a magnetic field might be resulted in to exist. Us delve into the causes of magnetic fields in succeeding chapters.

You are watching: Which of the following best describes magnetic field lines?

A magnetic field is a vector field. That is, it is an infinite set of vectors, one in ~ each allude in the an ar of space where the magnetic field exists. We use the expression “magnetic field” come designate both the infinite set of vectors, and, as soon as one is talking around the magnetic field at a allude in space, the one magnetic field vector in ~ that point in space. We use the prize $$\vecB$$ to represent the magnetic field. The most basic effect that a magnetic field is to exert a talk on things that has actually a property well-known as magnetic dipole moment, and, the finds chin in the magnetic field. A fragment or thing that has a non-zero worth of magnetic dipole minute is referred to as a magnetic dipole. A magnetic dipole is a bar magnet. The worth of the magnitude of the magnetic dipole minute of an item is a measure of how solid a bar magnet the is. A magnetic dipole has two ends, recognized as poles—a north pole and also a southern pole. Magnetic dipole moment is a building of matter which has direction. We can define the direction, the the magnetic dipole moment of an object, by considering the object to be an arrow whose north pole is the arrowhead and also whose south pole is the tail. The direction in which the arrowhead is pointing is the direction of the magnetic dipole minute of the object. The unit of magnetic dipole moment is the $$A\cdot m^2$$ (ampere meter-squared). While magnetic compass needles come in a range of magnetic dipole moments, a representative worth for the magnetic dipole minute of a compass needle is $$.1A\cdot m^2$$.

Again, the most basic effect that a magnetic field is to exert a speak on a magnetic dipole that finds chin in the magnetic field. The magnetic field vector, in ~ a given suggest in space, is the maximum possible torque-per-magnetic-dipole-moment-of-would-be-victim that the magnetic ar would/will exert on any kind of magnetic dipole (victim) that might find itself at the point in question. I need to say “maximum possible” since the talk exerted ~ above the magnetic dipole counts not just on the size of the magnetic ar at the suggest in room and the magnitude of the magnetic dipole minute of the victim, however it likewise depends on the orientation of the magnetic dipole family member to the direction the the magnetic ar vector. In fact:

\<\vec\tau=\vec\mu\times \vecB\label15-1\>

where:

$$\vec\tau$$ is the torque exerted on the magnetic dipole (the bar magnet) through the magnetic field,

$$\vec\mu$$ is the magnetic dipole moment of the magnetic dipole (the bar magnet, the victim), and

$$\vecB$$ is the magnetic field vector in ~ the ar in an are at i beg your pardon the magnetic dipole is.

For the cross product of any type of two vectors, the size of the cross product is the product of the magnitudes the the two vectors, time the sine of the edge the 2 vectors kind when placed tail come tail. In the instance of $$\vec\tau=\vec\mu\times \vecB$$, this means:

\<\tau=\mu\space B\cos \theta\>

In the SI mechanism of units, torque has units the $$N\cdot m$$ (newton-meters). For the systems on the right side that $$\tau=\mu\space B\cos \theta$$ to occupational out to it is in $$N\cdot m$$, what v $$\mu$$having systems of electrical dipole minute ($$A\cdot m^2$$ ) and $$\sin \theta$$ having actually no units at all, $$B$$ must have actually units that torque-per-magnetic-dipole-moment, namely, $$\fracN\cdot mA\cdot m^2$$. That combination unit is offered a name. That is called the tesla, abbreviated $$T$$.

\<1T= 1\fracN\cdot mA\cdot m^2\>

Consider a magnetic dipole having actually a magnetic dipole minute $$\mu=0.045 \, A\cdot m^2$$, oriented so the it provides an angle of $$23^\circ$$ through the direction the a uniform magnetic ar of magnitude $$5.0\times10^-5 T$$ as portrayed below. Uncover the torque exerted top top the magnetic dipole, by the magnetic field.

Recall the the arrowhead represents the north pole that the bar magnet that a magnetic dipole is. The direction of the torque is such the it often tends to cause the magnetic dipole to point in the direction that the magnetic field. Because that the case depicted above, that would be clockwise as viewed from the vantage point of the creator of the diagram. The size of the torque for such a situation can be calculated together follows:

\<\tau=\mu B\sin\theta\>

\<\tau=(.045A\cdot m^2)(5.0\times10^-5T)\sin 23^\circ\>

\<\tau=8.8\times 10^-7 A\cdot m^2 \cdot T\>

Recalling that a tesla is a $$\fracN\cdot mA\cdot m^2$$ we have:

\<\tau=8.8\times 10^-7 A\cdot m^2 \cdot \fracN\cdot mA\cdot m^2\>

\<\tau=8.8\times 10^-7 N\cdot m\>

Thus, the speak on the magnetic dipole is $$\tau=8.8\times 10^-7 N\cdot m$$ clockwise, as perceived from the vantage allude of the creator the the diagram.

A particle having a magnetic dipole minute $$\vec\mu=0.025 A\cdot m^2 \hati-0.035 A\cdot m^2 \hatj+0.015 A\cdot m^2 \hatk$$ is in ~ a suggest in space where the magnetic field $$\vecB=2.3 mT \hati+5.3mT\hatj-3.6mT\hatk$$. Discover the speak exerted on the bit by the magnetic field

\<\vec\tau=\vec\mu\times \vecB\>

\< \vec\tau=\beginvmatrix \hati&\hatj&\hatk\\ 0.025A\cdot m^2&-0.035 A\cdot m^2&0.015 A\cdot m^2 \\ 0.0023 \fracNmAm^2 &0.0053 \fracNmAm^2 & -0.0036 \fracNmAm^2\endvmatrix\>

\<\vec\tau=\hati \Big<(-0.035Am^2)(-0.0036 \fracNmAm^2)-(0.015Am^2)(0.0053 \fracNmAm^2) \Big>\>

\<+\hatj \Big< (0.015Am^2)(0.0023 \fracNmAm^2)-(0.025Am^2)(-0.0036 \fracNmAm^2) \Big>\>

\<+\hatk \Big< (0.025Am^2)(0.0053 \fracNmAm^2)-(-0.035Am^2)(0.0023 \fracNmAm^2) \Big>\>

\<\vec\tau=1.2\times10^-4 Nm \hati-1.2\times 10^-4 Nm \hatj+2.1\times 10^-4 Nm\hatk\>

## The Magnetic pressure Exerted upon a Magnetic Dipole

A uniform magnetic ar exerts no pressure on a bar magnet that is in the magnetic field. You should most likely pause below for a moment and let that sink in. A uniform magnetic field exerts no force on a bar magnet the is in the magnetic field.

You have probably had some experience with bar magnets. You recognize that choose poles repel and also unlike poles attract. And, native your research of the electric field, you have actually probably (correctly) hypothesized that in the field allude of view, the way we see this is that one bar magnet (call that the source magnet) creates a magnetic field in the region of an are around itself, and, that if there is an additional bar magnet in that region of space, it will certainly be influenced by the magnetic field it is in. We have already discussed the reality that the victim bar magnet will suffer a torque. However you know, from your endure with bar magnets, the it will likewise experience a force. How can that be once I just stated that a uniform magnetic field exerts no pressure on a bar magnet? Yes, of course. The magnetic ar of the resource magnet have to be non-uniform. Enough about the nature the the magnetic field of a bar magnet, I’m claimed to save that because that an upcoming chapter. Suffice it come say that it is non-uniform and to focus our attention on the result of a non-uniform ar on a bar magnet that finds chin in the magnetic field.

First the all, a non-uniform magnetic field will exert a torque on a magnetic dipole (a bar magnet) simply as before ($$\vec\tau=\vec\mu\times \vecB$$). But, a non-uniform magnetic ar (one because that which the magnitude, and/or direction, depends on position) additionally exerts a pressure on a magnetic dipole. The force is given by:

\<\vecF_B=\nabla (\vec\mu\cdot \vecB) \label15-2\>

where

$$\vecF_B$$ is the force exerted by the magnetic field $$\vecB$$ on a particle having a magnetic dipole moment $$\vec\tau$$ $$\vec\mu$$ is the magnetic dipole of the "victim", and, $$\vecB$$ is the magnetic field at the place in an are where the victim find itself. To evaluate the force, as soon as must recognize $$\vecB$$ together a role of $$x,y$$ and $$z$$ (wherea $$\vec\mu$$ is a constant) .

Note the after you take the gradient the $$\vec\mu\cdot \vecB$$, you need to evaluate the an outcome at the values of $$x,y$$ and $$z$$ matching to the ar of the victim.

See more: How To Use Ziggurat Civ 6 ), Civilization 6: Guide To Winning With Sumeria

Just come make sure that girlfriend know how to use this equation, please keep in mind that if $$\vec\mu$$ and $$\vecB$$ are expressed in $$\hati,\hatj,\hatk$$ notation, so the they appear as $$\vec\mu=\mu_x \hati+\mu_y \hatj+\mu_z \hatk$$ and also $$\vecB=B_x \hati+B_y \hatj+B_z \hatk$$ respectively, then:

\<\vec\mu\cdot\vecB=(\vec\mu=\mu_x \hati+\mu_y \hatj+\mu_z \hatk)(\vecB=B_x \hati+B_y \hatj+B_z \hatk)\>

\<\vec\mu\cdot\vecB=\mu_x B_x+\mu_y B_y+\mu_z B_z\>

And the gradient of $$\vec\mu\cdot\vecB$$ (which through equation $$\ref15-2$$ is the pressure we seek) is offered by

\<\nabla (\vec\mu\cdot \vecB)=\frac\partial (\vec\mu\cdot \vecB)\partial x \hati +\frac\partial (\vec\mu\cdot \vecB)\partial y \hatj+\frac\partial (\vec\mu\cdot \vecB)\partial z \hatk\>

where derivatives in this equation can (using $$\vec\mu\cdot\vecB=\mu_x B_x+\mu_y B_y+\mu_z B_z$$ from just above) have the right to be to express as:

\<\frac\partial (\vec\mu\cdot\vecB)\partial x=\mu_x \frac\partial B_x\partial x+\mu_y \frac\partial B_y\partial x+\mu_z \frac\partial B_z\partial x\>

\<\frac\partial (\vec\mu\cdot\vecB)\partial y=\mu_x \frac\partial B_x\partial y+\mu_y \frac\partial B_y\partial y+\mu_z \frac\partial B_z\partial y\>

\<\frac\partial (\vec\mu\cdot\vecB)\partial z=\mu_x \frac\partial B_x\partial z+\mu_y \frac\partial B_y\partial z+\mu_z \frac\partial B_z\partial z\>

where we have actually taken benefit of the fact that the materials of the magnetic dipole moment of the victim room not functions of position. Likewise note that the derivatives space all partial derivatives. Partial derivatives are the basic kind in the sense that, when, for instance, you take the derivative through respect come $$x$$, you space to law $$y$$ and also $$z$$ together if they to be constants. Finally, the is necessary to realize that, after you take it the derivatives, you need to plug the values of $$x,y$$ and also $$z$$ corresponding to the place of the magnetic dipole (the victim), into the provided expression because that the force.